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Asked in AIPMT 2001 · Circuits with diodes
Idea: settle which diode conducts first, then add only the resistances lying on the path that actually carries current.
The cell drives current one way round the loop. D₁ is oriented against that direction, so its branch behaves as an open circuit and carries nothing.
D₂ is oriented with it, so the current leaves the cell's branch and returns through D₂ and the 30 Ω beside it.
On that path the cell's own branch contributes 20 Ω and the conducting branch contributes 30 Ω.
Total resistance: 20+30=50 Ω.
I=5/(50) A, which is 0.1 A.
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